Target Exam

CUET

Subject

General Aptitude Test

Chapter

Quantitative Reasoning

Topic

Permutation & Combination

Question:

In how many different ways can the letters of the word 'OFFICE' be arranged so that the vowels never come together?

Options:

288

216

576

192

Correct Answer:

288

Explanation:

The correct answer is Option (1) → 288

1. Total number of arrangements of 'OFFICE':

The word 'OFFICE' has 6 letters: O, F, F, I, C, E.

  • Total letters = 6
  • Repeated letter = F (appears 2 times)

$\text{Total arrangements} = \frac{6!}{2!} = \frac{720}{2} = 360$

2. Number of arrangements where all vowels are together:

The vowels are O, I, E. We treat them as a single unit or "block": (OIE).

  • The units to arrange are now: (OIE), F, F, C.
  • Total units = 4.
  • Arrangements of these 4 units (with two F's): $\frac{4!}{2!} = \frac{24}{2} = 12$.
  • Arrangements of the vowels within the block (O, I, E): $3! = 6$.

$\text{Arrangements with vowels together} = 12 \times 6 = 72$

3. Number of arrangements where vowels are never together:

$\text{Vowels never together} = \text{Total arrangements} - \text{Arrangements with vowels together}$

$\text{Result} = 360 - 72 = 288$